CPSC 490 Problem Solving in Computer Science Dijkstra, Floyd-Warshall, Union-Find, and Toposort Lucca Siaudzionis and Jack Spalding-Jamieson 2020/01/14 University of British Columbia
Announcements • Assignment 1 has been released. • After today’s lecture, you will be able to solve most of the problems. • If you haven’t started yet, please try to do so today. • If anyone still does not have an account, please request one on piazza ASAP. 1
The Shortest Path Problem for Weighted Graphs: Given Pair Input : A graph with non-negative edge-weights, and labelled nodes u and v . Output : The shortest distance from u to v . What if we can solve a more general problem? 2
The Shortest Path Problem for Weighted Graphs: Single Source Input : A graph with non-negative edge-weights, and a labelled node u . Output : The distance from u to each other node in the graph. To solve the original problem: Output the distance from u to v after running the solution here. Ideas for solving this problem: This looks very similar to our use-case for BFS, but now with edge-weights. Maybe we can do something similar? 3
Dijkstra’s Algorithm Graph traversal algorithm which starts at a given node, and explores the other nodes by order of their distance to the given node. • Time complexity: O ( m + n log m ) Things to consider for a Dijkstra problem: • What needs to be done when ... • Visiting a new node? • Visiting an already-visited node? • Visiting the target node? • Do we need to modify the graph in any way? • Do we need to create and connect copies of the graph? 4
The Shortest Path Problem for Weighted Graphs: Example (1) 12 2 9 4 5 22 3 3 4 Our source node is the purple node in the top-left. We can use Dijkstra’s algorithm to solve the single-source shortest path problem on this graph. 5
The Shortest Path Problem for Weighted Graphs: Example (2) ∞ 0 12 2 9 ∞ ∞ 4 ∞ 5 22 3 ∞ ∞ 3 4 ∞ Initialize the distance of the source vertex to be 0, and all non-source distances to be ∞ . 6
The Shortest Path Problem for Weighted Graphs: Example (3) ∞ 0 12 2 9 ∞ ∞ 4 12 5 22 3 ∞ ∞ 3 4 ∞ Find the closest node to the source. 7
The Shortest Path Problem for Weighted Graphs: Example (4) 14 0 12 2 9 ∞ ∞ 4 12 5 22 3 ∞ ∞ 3 4 ∞ Find the second-closest node to the source. 8
The Shortest Path Problem for Weighted Graphs: Example (5) 14 0 12 2 9 ∞ ∞ 4 12 5 22 3 ∞ 19 3 4 ∞ 9
The Shortest Path Problem for Weighted Graphs: Example (6) 14 0 12 2 9 22 ∞ 4 12 5 22 3 ∞ 19 3 4 ∞ 10
The Shortest Path Problem for Weighted Graphs: Example (7) 14 0 12 2 9 22 ∞ 4 12 5 22 3 ∞ 19 3 4 23 11
The Shortest Path Problem for Weighted Graphs: Example (8) 14 0 12 2 9 22 26 4 12 5 22 3 26 19 3 4 23 12
Graph Traversals: Dijkstra’s Algorithm Implementation vector<int> dijkstra(vector<vector<int>> &adj, 1 vector<vector<int>> &w, int s) { 2 vector<bool> visited(adj.size()); 3 vector<int> dist(n); // shortest distance from s 4 // pair: first is distance (sorted by), second is node 5 priority_queue<pair<int,int>, vector<pair<int,int>>, 6 greater<pair<int,int>>> q; q.push({0, s}); 7 while (!q.empty()) { 8 int cur_dist, cur_node; 9 tie(cur_dist,cur_node) = q.top(); q.pop(); 10 if (visited[cur_node]) continue; visited[cur_node] = true; 11 dist[cur_node] = cur_dist; 12 for (int i = 0; i < adj[cur_node].size(); ++i) 13 q.emplace(dist[cur_node] + w[i], adj[i]); 14 } 15 13 } 16
Discussion Problem 1 - Odd Path Input: A graph G with non-negative weighted directed edges, a node s , and a node e . Output: The shortest path from s to e that contains an odd number of edges. 14
Discussion Problem 1 - Insight Duplicate the graph • Every node u becomes nodes u 1 and u 2 • Every edge u → v becomes edges u 1 → v 2 and u 2 → v 1 Paths from u 1 to v 1 and from u 2 to v 2 contain an even number of edges. Paths from u 1 to v 2 and from u 2 to v 1 contain an odd number of edges. Run Dijkstra to find the shortest path from s 1 to e 2 . 15
All pairs Shortest Paths: Problem Statement Input : A graph with edge-weights. The edge-weights might be negative, but there will be no negative cycles. Output : For each pair u , v of vertices, the length of the smallest path between u and v (possibly ∞ ). Observation 1: If the edge-weights were non-negative, we could use Dijkstra to solve this in O ( nm log n ) time. Observation 2: Similarly to BFS and Dijkstra, any shortest path between u and v using more than two edges uses a shortest path of some w ∈ N ( u ) and v . 16
All pairs Shortest Paths: Floyd-Warshall initialize a matrix dist [][] to infinity 1 set dist[u][u] to 0 for all nodes u 2 set dist[u][v] to w(u,v) for all edges e = (u,v) 3 for k = 1 to |V|: 4 for i = 1 to |V|: 5 for j = 1 to |V|: 6 dist[i][j] = min(dist[i][j], 7 dist[i][k] + dist[k][j]) 8 Time Complexity: O ( V 3 ). Runs very fast in practice because of tightly nested loops. This is one of the cases where you do want to use an adjacency matrix! 17
Discussion Problem 2 - Maximin Lucca is in a country with N cities and M roads connecting them. The country is full of overhead bridges, so each road connecting cities u i and v i has a maximum height h i allowed for all vehicles in it. Lucca likes to rent very tall trucks, and was wondering what is the tallest truck that he can rent to go from city S to city E . 18
Discussion Problem 2 - Insight Change the step that joins distances to use a min instead of a sum . Change the step that compares distances to use a max instead of a min . Example in Floyd-Warshall: dist[i][j] = max(dist[i][j], min(dist[i][k], dist[k][j])) 1 19
Union-Find (Disjoint Set Union) Union-Find is a data structure used to keep track of disjoint sets. It allows for two operations: • Union : Join two sets. • Find : Find the ”representative” of an element’s set. Two elements are in the same set ⇒ they have the same representative (i.e. find(a) == find(b) ). ⇐ Time Complexity: O ( α ( n )) ← inverse Ackermann function Essentially constant time ( O (1)) for any practical purposes! 20
Union-Find Example: Initial state 1 2 3 F=2 F=3 F=1 5 4 6 F=4 F=5 F=6 21
Union-Find Example: Union(1, 2) 1 2 3 F=1 F=3 F=1 4 5 6 F=5 F=4 F=6 22
Union-Find Example: Union(4, 5) 1 2 3 F=1 F=3 F=1 4 5 6 F=5 F=4 F=6 23
Union-Find Example: Union(3, 6) 1 2 3 F=6 F=1 F=1 5 4 6 F=4 F=4 F=6 24
Union-Find Example: Union(2, 3) 1 2 3 F=1 F=1 F=1 4 5 6 F=4 F=4 F=1 25
Union-Find: Find Function vector<int> parent; 1 // in an initial state, parent[i] = i for every i 2 3 int find(int x) { 4 if (parent[x] != x) { 5 parent[x] = find(parent[x]); // path-compression 6 } 7 return parent[x]; 8 } 9 26
Union-Find: Union by Rank (Join) Function vector<int> parent; 1 vector<int> rnk; // rank -- initialized to all 0s 2 3 void join(int a, int b) { 4 // we are only interested in the representatives 5 a = find(a); b = find(b); 6 if (a == b) return; // already joined 7 8 if (rank[a] > rank[b]) parent[b] = a; 9 else if (rank[b] > rank[a]) parent[a] = b; 10 else { 11 parent[b] = a; 12 rank[a]++; 13 } 14 } 15 27
Topological Sort Remember this graph? CPSC 110 CPSC 121 CPSC 210 CPSC 213 CPSC 221 CPSC 310 CPSC 313 CPSC 320 CPSC 410 CPSC 415 CPSC 420 Pictured: Graph of the UBC Computer Science pre-requisite chain. If someone asked you for a correct sequence to take all these courses, how would you compute such sequence? 28
Topological Sort Problem formulation: • You have N tasks (numbered 1 to N ). • You have M requirements ( u , v ), meaning that task u must be completed before task v . Output: A valid sequence to complete all tasks. 29
Topological Sort Problem graph: • Every task can be represented by a node in the graph. • Every requirement ( u , v ) can be represented by a directed edge from u to v . Note that the resulting graph is a DAG iff the problem has a solution. 30
Topological Sort - Intuition Can you start with a task with in-degree 1 greater than 0? • No. If an edge arrives at a task, then the task on the other side must be completed first. • That means that we can only start with tasks with in-degree equal to 0. 1 in-degree of a node is the number of edges that arrive at that node. 31
Topological Sort - Possible Starting Nodes 32
Topological Sort - Intuition What happens after we complete a task? • That means that all the tasks that depended on it no longer need to await its completion. • Which means that we can essentially delete that node and all its edges from the graph. • Any problem with this feature is called self-reducible . Later in this course, we will discuss another important self-reducible problem. 33
Topological Sort - Completing a Task 34
Topological Sort - Completing a Task 35
Topological Sort - Intuition What do we do now? Repeat until all tasks are completed! 36
Topological Sort - Possible Starting Nodes 37
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